Verified answer

Ответ:

х/у=(-3+-sqrt(5))/2

Объяснение:

Ответ:      

[tex]\left\{\begin{array}{l}x+y=3xy\\x^2+y^2=\dfrac{1}{3}\end{array}\right\ \ \left\{\begin{array}{l}(x+y)^2=9(xy)^2\\x^2+y^2=\dfrac{1}{3}\end{array}\right\ \ \left\{\begin{array}{l}x^2+y^2+2xy=9(xy)^2\\x^2+y^2=\dfrac{1}{3}\end{array}\right\\\\\\\left\{\begin{array}{l}\dfrac{1}{3}+2xy=9(xy)^2\\x^2+y^2=\dfrac{1}{3}\end{array}\right\ \ \left\{\begin{array}{l}9(xy)^2-2(xy)-\dfrac{1}{3}=0\\x^2+y^2=\dfrac{1}{3}\end{array}\right\ \ \left\{\begin{array}{l}27(xy)^2-6(xy)-1=0\\x^2+y^2=\dfrac{1}{3}\end{array}\right[/tex]

[tex]\star \ \ t=xy\ ,\ 27t^2-6t-1=0\ ,\ \ D/4=9+27=36\ ,\\\\t_1=\dfrac{3-6}{27}=-\dfrac{1}{9}\ \ ,\ \ \ t_2=\dfrac{3+6}{27}=\dfrac{1}{3}\ \ \Rightarrow \ \ \ (xy)_1=-\dfrac{1}{9}\ \ ,\ \ (xy)_2=\dfrac{1}{3}\ \ \star[/tex]  

[tex]a)\ \ \left\{\begin{array}{l}x+y=3xy\\xy=-\dfrac{1}{9}\end{array}\right\ \ \left\{\begin{array}{l}x+y=-\dfrac{1}{3}\\xy=-\dfrac{1}{9}\end{array}\right\ \ \left\{\begin{array}{l}y=-x-\dfrac{1}{3}\\x(-x-\dfrac{1}{3} )=-\dfrac{1}{9}\end{array}\right\\\\\\-x^2-\dfrac{1}{3}\, x+\dfrac{1}{9}=0\ \ \Big|\cdot (-9)\ \ \ \Rightarrow \ \ \ 9x^2+3x-1=0\ \ ,\ \ D=9+36=45\ \ ,[/tex]

[tex]x_1=\dfrac{-3-3\sqrt5}{2\cdot 9}=\dfrac{-1-\sqrt5}{6}\ \ ,\ \ \ x_2=\dfrac{-1+\sqrt5}{6}\\\\\\y_1=-x-\dfrac{1}{3}=\dfrac{1+\sqrt5}{6}-\dfrac{1}{3}=\dfrac{\sqrt5-1}{6}\ \ ,\\\\\\y_2=-x-\dfrac{1}{3}=\dfrac{1-\sqrt5}{6}-\dfrac{1}{3}=\dfrac{-\sqrt5-1}{6}\\\\\\\boldsymbol{\dfrac{x_1}{y_1}}=\dfrac{-1-\sqrt5}{\sqrt5-1}=\dfrac{-(1+\sqrt5)^2}{(\sqrt5-1)(\sqrt5+1)}=\dfrac{-(6+2\sqrt5)}{5-1}=\bf \dfrac{-3-\sqrt5}{2}[/tex]

[tex]\boldsymbol{\dfrac{x_2}{y_2}}=\dfrac{-1+\sqrt5}{-\sqrt5-1}=\dfrac{(\sqrt5-1)^2}{-(\sqrt5+1)(\sqrt5-1)}=\dfrac{6-2\sqrt5}{-(5-1)}=\dfrac{2(3-\sqrt5)}{-4}=\bf \dfrac{-3+\sqrt5}{2}[/tex]  

[tex]b)\ \ \left\{\begin{array}{l}x+y=3xy\\xy=\dfrac{1}{3}\end{array}\right\ \ \left\{\begin{array}{l}x+y=1\\xy=\dfrac{1}{3}\end{array}\right\ \ \left\{\begin{array}{l}y=1-x\\x(1-x)=\dfrac{1}{3}\end{array}\right\ \ \left\{\begin{array}{l}x+y=1\\x-x^2-\dfrac{1}{3}=0\end{array}\right\\\\\\-x^2+x-\dfrac{1}{3}=0\ \ \Big|\cdot (-3)\ \ \ \Rightarrow \ \ \ 3x^2-3x+1=0\ \ ,\ \ D=9-4\cdot 3=-3 < 0[/tex]

Так как  D<0 , то действительных корней нет . Система не имеет решений .

Ответ:     [tex]\boldsymbol{\dfrac{x_1}{y_1}=\dfrac{-3-\sqrt5}{2}}\ \ ,\ \ \boldsymbol{\dfrac{x_2}{y_2}=\dfrac{-3+\sqrt5}{2}}[/tex]   .  

 P.S.  Если системы решать , введя замену переменной    [tex]\dfrac{x}{y}=t[/tex]  ,  то сразу можно получить значения дроби .

[tex]a)\ \ \dfrac{x}{y}=t\ \ ,\ \ x=ty\ \ ,\ \ \ \left\{\begin{array}{l}ty+y=-\dfrac{1}{3}\\ty\cdot y=-\dfrac{1}{9}\end{array}\right\ \ \left\{\begin{array}{l}y=-\dfrac{1}{3(t+1)}\\\dfrac{t}{9(t+1)^2}=-\dfrac{1}{9}\end{array}\right\\\\\\\left\{\begin{array}{l}t=\dfrac{x}{y}\\t^2+2t+1=-t\end{array}\right\ \ \ \Rightarrow \ \ \ t^2+3t+1=0\ ,\ t_{1,2}=\dfrac{-3\pm \sqrt5}{2}\ \ \Rightarrow[/tex]  

[tex]\dfrac{x}{y}=\dfrac{-3-\sqrt5}{2}\ \ \ ili\ \ \ \dfrac{x}{y}=\dfrac{-3+\sqrt5}{2}[/tex]  

[tex]b)\ \ \dfrac{x}{y}=t\ ,\ \ x=ty\ \ ,\ \left\{\begin{array}{l}x+y=1\\xy=\dfrac{1}{3}\end{array}\right\ \ \left\{\begin{array}{l}ty+y=1\\ty\cdot y=\dfrac{1}{3}\end{array}\right\ \ \left\{\begin{array}{l}y=\dfrac{1}{t+1}\\\dfrac{t}{(t+1)^2}=\dfrac{1}{3}\end{array}\right\\\\\\t^2+2t+1=3t\ \ \ \Rightarrow \ \ \ t^2-t+1=0\ \ ,\ \ D=1-4=-3 < 0\ ,\ t\in \varnothing[/tex]